Question

In triangle ABC, AD is perpendicular to BC and BD is equal to 1/3×CD. Prove that 2CA^2=2 AB^2 + BC^2.​

Answer 1

Answer:

In right DADB, AB2= AD2+ BD2[By Pythagoras theorem] – (1)

In right DADC, AC2= AD2+ CD2[By Pythagoras theorem] – (2)

Subtract (1) from (2), we get

AC2– AB2= AD2+ CD2– AD2– BD2

AC2– AB2= CD2– BD2--- (3)

Given BD =1/3 CD

From the figure, BD + DC = BC

⇒ (1/3)CD + CD = BC

⇒ (4/3)CD = BC

∴ CD = 3BC/4

⇒ BD = BC/4

Equation (3) becomes,

AC2– AB2= (3BC/4)2– (BC/4)2

=(8/16)BC2

⇒ AC2– AB2= (1/2) BC2

⇒ 2AC2– 2AB2= BC2

⇒ 2AC2= 2AB2+ BC2